\newproblem{lay:7_2_23}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.2.23}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Consider the quadratic form $Q(\mathbf{x})=\mathbf{x}^TA\mathbf{x}$ when $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\det\{A\}\neq 0$. 
	If $\lambda_1$ and $\lambda_2$ are the eigenvalues of $A$, then the characteristic polynomial of $A$ can be written in two ways: $\det\{A-\lambda I\}$ and 
	$(\lambda-\lambda_1)(\lambda-\lambda_2)$. Use this fact to show that $\lambda_1+\lambda_2=a+d$ (the diagonal entries of $A$) and $\lambda_1\lambda_2=\det\{A\}$.
}{
   % Solution
	We may express the characteristic polynomial as
	\begin{center}
		$\begin{array}{rcl}P(\lambda)&=&\det\{A-\lambda I\}=\det\left|\begin{array}{cc}a-\lambda & b \\ c & d-\lambda\end{array}\right|\\
		  &=&(a-\lambda)(d-\lambda)-bc\\
			&=&\lambda^2-(a+d)\lambda+(ad-bc)\\
			&=&\lambda^2-\mathrm{Trace}\{A\}\lambda+\det\{A\} \\
		 P(\lambda)&=&(\lambda-\lambda_1)(\lambda-\lambda_2)=\lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2\end{array}$
	\end{center}
	Identifying coefficients we see that
	\begin{center}
		$\mathrm{Trace}\{A\}=a+d=\lambda_1+\lambda_2$ \\
		$\mathrm{det}\{A\}=ad-bc=\lambda_1\lambda_2$ \\
	\end{center}
}
\useproblem{lay:7_2_23}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

